CHAPTER 15 REACTIONS AND EQUILIBRIA INVOLVING ACIDS, BASES, AND SALTS

1
CHAPTER 15REACTIONS AND EQUILIBRIA INVOLVING
ACIDS, BASES, AND SALTS
2
Common ion effect- The addition of an ion
already present(common) in a system causes
equilibrium to shift away from the common ion.

3
For example, the addition of concentrated HCl to
a saturated solution of NaCl will cause some
solid NaCl to precipitate out of solution. The
NaCl has become less soluble because of the
addition of additional chloride ion. This can
be explained by the use of LeChatelier's
Principle. NaCl(s) ? Na Cl-
4
The addition of a common ion to a solution of a
weak acid makes the solution less acidic.
HC2H3O2 ? H C2H3O2-If we add NaC2H3O2,
equilibrium shifts to undissociated HC2H3O2,
raising pH. The new pH can be calculated by
putting the concentration of the anion into the
Ka equation and solving for the new H.
5
Buffered solution- A solution that resists
changes in pH when hydroxide ions or protons are
added. A buffer solution usually consists of a
solution of a weak acid and its salt or a weak
base and its salt.
6
Ex. HC2H3O2/C2H3O2- buffer systemAddition of
strong acid H C2H3O2- ? HC2H3O2Addition
of strong base OH- HC2H3O2 ?H2O C2H3O2-
7
NH3/NH4 buffer system Addition of strong
acid H NH3 ? NH4Addition of strong base
OH- NH4 ? NH3 H2O
8
Buffer capacity- The amount of acid or base that
can be absorbed by a buffer system without a
significant change in pH. In order to have a
large buffer capacity, a solution should have
large concentrations of both buffer
components.
9
One way to calculate the pH of a buffer
system is with the Henderson-Hasselbach
equation. pH pKa log base
acidpH pKa log A-
HA
10
For a particular buffering system all solutions
that have the same ratio of A-/HA have the
same pH. Optimum buffering occurs when HA
A- and the pKa of the weak acid used should be
as close as possible to the desired pH of the
buffer system.
11
The Henderson-Hasselbach (HH) equation needs to
be used cautiously. It is sometimes used as a
quick, easy equation to plug numbers into. A Ka
or Kb problem requires a greater understanding of
the factors involved and can always be used
instead of the HH equation.
12
Hints for Solving Buffer Problems1. Determine
major species involved initially.2. If chemical
reaction occurs, write equation and solve
stoichiometry in moles, then change to
molarity.3. Write equilibrium equation.4. Set
up equilibrium expression (Ka or Kb) or HH
equation.5. Solve.6. Check logic of answer.
13
Ex. A solution is 0.120 M in acetic acid and
0.0900 M in sodium acetate. Calculate the H
at equilibrium. The Ka of acetic acid is 1.8 x
10-5.
WA CB Ka!

• Rxn HC2H3O2 ? H C2H3O2-
• Initial 0.120 0 0.0900
• Change -x x x
• Equil. 0.120-x x 0.0900 x
• Ka x (0.0900 x) ? x (0.0900) 1.8 x
  10-5 0.120-x
  0.120
• x 2.4 x 10-5 M
• H 2.4 x 10-5

14
Using the Henderson-Hasselbach equationpKa
-log 1.8 x 10-5 4.74pH 4.74 log
(0.0900/0.120) 4.62H antilog (-4.62)
2.4 x 10-5
15
Ex. Calculate the pH of the above buffer system
when 100.0 mL of 0.100 M HCl is added to 455 mL
of solution.

• 0.100 L HCl x 0.100 M 0.0100 mol H
• 0.455 L C2H3O2- x 0.0900 M 0.0410 mol C2H3O2-
• 0.455 L HC2H3O2 x 0.120 M 0.0546 mol HC2H3O2
  H C2H3O2- ?
  HC2H3O2
• Before 0.0100 mol 0.0410 mol 0.0546
  mol
• Change -0.0100 mol -0.0100 mol 0.0100 mol
• After 0 0.0310 mol
  0.0646 mol
• 0.0310 mol acetate / 0.555 L solution 0.0559 M
  acetate
• 0.0646 mol acetic acid/0.555 L solution 0.116 M
  acetic acid

WA CB Ka!
16

• Rxn HC2H3O2 ? H C2H3O2
• Initial 0.116 M 0 0.0559 M
• Change -x x x
• Equil. 0.116-x x 0.0559
  xKa 1.8 x 10-5 x(0.0559x) ? x(0.0559)
• 0.116-x
  0.116
• x 3.74 x 10-5 M H
• pH 4.43

17
Acid-Base Titrations
18
titrant-solution of known concentration (in
buret) The titrant is added to a solution of
unknown concentration until the substance being
analyzed is just consumed (stoichiometric point
or equivalence point).
19
pH or titration curve -plot of pH as a function
of the amount of titrant added.
20
Types of Acid-Base Titrations
21
1. Strong acid-strong baseSimple reaction H
OH- ? H2O The pH is easy to calculate because
all reactions go to completion. At the
equivalence point, the solution is neutral.
22
The pH Curve for the Titration of 50.0 mL of
0.200 M HNO3 with 0.100 M NaOH
Strong base left
Only neutral salt
Only strong acid left
23
Ex. 100.0 mL of 1.00 M HCl is titrated with
0.500 M NaOH. Calculate the H after 50.0 mL of
base has been added.

• 0.1000L x 1.00 M 0.100 mol H
• 0.0500L x 0.500 M 0.0250 mol OH
• H OH- ?
  H2O
• Before 0.100 mol 0.0250 mol 0
  
• Change -0.0250 -0.0250 0.0250
• After 0.0750 mol 0
  0.0250 mol
• 0.0750 mol H/ (0.100L 0.0500 L)
• 0.500 M H

24
Calculate the H after 200 mL of base has been
added.

• 0.200L x 0.500 M 0.100 mol OH-
• H OH- ?
  H2O
• Before 0.100 mol 0.100 mol 0
• Change -0.100 -0.100 0.100
• After 0 0
  0.100 mol
• H is not zero. The H of pure water is
• 1.0 x 10-7, therefore pH 7 H 1.0 x 10-7

25
Calculate the pH after 300 mL of base has been
added.

• 0.300L x 0.500 M 0.150 mol OH-
• H OH- ? H2O
• Before 0.100 mol 0.150 mol ---
• Change -0.100 -0.100 ---
• After 0 0.050 mol
  ---
• OH- 0.050 mol/0.400L 0.125 M OH-
• pOH 0.913 pH 13.097

26
2. Weak acid - strong baseThe reaction of a
strong base with a weak acid is assumed to go to
completion. Before the equivalence point, the
concentration of weak acid remaining and the
conjugate base formed are determined. At halfway
to the equivalence point, pH pKa.
27
The pH Curve for the Titration of 50.0 mL of
0.100 M HC2H3O2 with 0.100 M NaOH
SBgtWA Stoich pOH -log SB
SB WA Basic salt Kb
WAgtSB HAA- ½ eq pt pH pKa
WA gtSB Stoich Ka
WA only Ka
WA gt SB Stoich Ka
28
At the equivalence point, a basic salt is present
and the pH will be greater than 7. After the
equivalence point, the strong base will be the
dominant species and a simple pH calculation can
be made after the stoichiometry is done.
29
(No Transcript)
30
Ex. 30.0 mL of 0.10 M NaOH is added to 50.0 mL
of 0.10 M HF. (Ka of HF 7.2 x 10-4) Determine
the pH of the final solution.

• 0.0300L x 0.10 M 0.00300 mol OH-
• 0.0500L x 0.10 M 0.00500 mol HF
• Stoichiometry OH-
  HF ? H2O F-
• Before 0.00300 mol 0.00500 mol ---
  0
• Change -0.003 -0.003
  0.003
• After 0 0.00200 mol
  --- 0.00300 mol
• 0.00200 mol/(0.030L 0.050L) 0.0250M HF
• 0.00300 mol/(0.030L 0.050L) 0.0375M F-

WA CB Ka!
31
Equilibrium

• Rxn HF ? H F-
• Initial 0.025 0 0.0375
• Change -x x x
• Equil. 0.025-x x 0.0375 x
• Ka 7.2 x 10-4 x (0.0375 x) ? x(0.0375)
• 0.0250-x
  0.0250
• x 4.8 x 10-4
• H 4.8 x 10-4
• pH 3.32

32
Ex. 50.0 mL of 0.10 M NaOH is added to 50.0 mL
of 0.10 M HF. (Ka of HF 7.2 x 10-4) Determine
the pH of the final solution.

• 0.050L x 0.10 M 0.0050 mol OH-
• 0.050L x 0.10 M 0.0050 mol HFStoichiometry
• OH- HF ? H2O
  F-
• Before 0.0050 mol 0.0050 mol ----
  0
• After 0 0
  ----- 0.0050 mol
• 0.0050 mol/(0.050L 0.050L) 0.050M F-

CB only Kb!
33
Equilibrium

• Rxn F- H2O ? HF OH-
• Initial 0.050 ---- 0
  0
• Change -x ----- x
  x
• Equil. 0.050-x ---- x
  x

34
Kb for F- 1.0 x 10-14/Ka for HF

• Kb 1.4 x 10-11 HFOH- x2 ?
  x2 F-
  0.050-x 0.050
• x 8.4 x 10-7 M
• OH- 8.4 x 10-7
• pOH 6.08
• pH 7.92

35
Ex. 60.0 mL of 0.10 M NaOH is added to 50.0
mL of 0.10 M HF. (Ka of HF 7.2 x 10-4)
Determine the pH of the final solution.

• 0.0600L x 0.10 M OH- 0.0060 mol OH-
• 0.0500L x 0.10 M HF 0.0050 mol HF
• Stoichiometry
• OH- HF ?
  H2O F-
• Before 0.0060 mol 0.0050 mol ------
  0
• After 0.0010 mol 0
  -------- 0.0050 mol
• OH- 0.0010 mol/0.110 L 9.09 x 10-3 M
• Strong base weak base (ignore weak base!)
• pOH 2.04 pH 11.96

36
Weak base - Strong acid Before the
equivalence point, a weak base equilibria exists.
Calculate the stoichiometry and then the weak
base equilibria. At the equivalence point, an
acidic salt is present and the pH is below 7.
After the equivalence point, the strong acid is
the dominant species. Use the H to find the
pH.
37
The pH Curve for the Titration of 100.0 mL of
0.050 M NH3 with 0.10 M HCI
SAltWB B HB ½ eq pt pOH pKb
Weak base only Kb
SA lt WB Stoich Kb
SA lt WB Stoich Kb
SAgt WB Stoich, pH -log SA
WB SA Acidic salt Ka
38
Ex. Calculate the pH when 100.0 mL of 0.050 M
NH3 is titrated with 10.0 mL of 0.10 M HCl. Kb
of NH3 1.8 x 10-5

• 0.100L x 0.050 M 0.0050 mol NH3
• 0.010L x 0.10 M 0.0010 mol H
• NH3 H ?
  NH4
• Before 0.0050 mol 0.0010 mol 0
• Change -0.0010 -0.0010 0.0010
• After 0.0040 mol 0 0.0010 mol
• 0.0010 mol/0.110 L 9.09 x 10-3M NH4
• 0.0040 mol/0.110 L 3.64 x 10-2M NH3

WB CA Kb!
39
Equilibrium

• Rxn NH3 H2O ? NH4 OH-
• Initial 0.0364 ----- 0.00909
  0
• Change -x x
  x
• Equil 0.0364-x 0.00909x
  x
• Kb 1.8 x 10-5 (0.00909x)x ? (0.00909)x
  0.0364-x 0.0364
• x OH- 7.21 x 10-5 pOH 4.14
• pH 9.86

40
Ex. Calculate the pH when 100.0 mL of 0.050 M
NH3 is titrated with 50.0 mL of 0.10 M HCl. Kb
of NH3 1.8 x 10-5

• 0.100L x 0.050 M 0.0050 mol NH3
• 0.050L x 0.10 M 0.0050 mol H
• NH3 H ?
  NH4
• Before 0.0050 mol 0.0050 mol 0
• Change -0.0050 -0.0050 0.0050
• After 0 0
  0.0050 mol0.0050 mol/ 0.150L 0.0333M NH4

CA only Ka!
41
Equilibrium

• Rxn NH4 H2O ? NH3 H3O
• Initial 0.0333 ----- 0
  0
• Change -x x
  x
• Equil. 0.0333-x x
  x
• Ka for NH4 1.0 x 10-14/Kb for NH3 5.56 x
  10-10
• 5.56 x 10-10 NH3H3O x2 ?
  x2 NH4 0.0333-x
  0.0333
• x 4.30 x 10-6 H pH 5.37

42
Ex. Calculate the pH when 100.0 mL of 0.050 M
NH3 is titrated with 60.0 mL of 0.10 M HCl.

• 0.100L x 0.050 M 0.0050 mol NH3
• 0.060L x 0.10 M 0.0060 mol H NH3
  H ? NH4
• Before 0.0050 mol 0.0060 mol 0
• Change -0.0050 -0.0050
  0.0050
• After 0 0.0010 mol
  0.0050 mol
• 0.0010 mol/0.160L 6.25 x 10-3 M H3O
• pH 2.20

Strong acid and CA Ignore CA(weak)!
43
Acid-Base Indicatorsend point- point in
titration where indicator changes color
44
When choosing an indicator, we want the
indicator end point and the titration equivalence
point to be as close as possible. Since strong
acid-strong base titrations have a large vertical
area, color changes will be sharp and a wide
range of indicators can be used. For titrations
involving weak acids or weak bases, we must be
more careful in our choice of indicator.
45
Indicators are usually weak acids, HIn. They
have one color in their acidic (HIn) form and
another color in their basic (In-) form.
A very common indicator, phenolphthalein, is
colorless in its HIn form and pink in its In-
form. It changes color in the range of pH 8-10.

46
(No Transcript)
47
Usually 1/10 of the initial form of the indicator
must be changed to the other form before a new
color is apparent.
48
The following equations can be used to
determine the pH at which an indicator will
change colorFor titration of an acidpH
pKa log 1/10 pKa-1For titration of a
basepH pKa log 10/1 pKa1
49
The useful range of an indicator is usually its
pKa 1. When choosing an indicator, determine
the pH at the equivalence point of the titration
and then choose an indicator with a pKa close to
that.
50
(No Transcript)
51
The pH Curve for the Titration of 100.0 mL of
0.10 M HCI with 0.10 M NaOH
52
The pH Curve for the Titration of 50 mL of 0.1 M
HC2H3O2 with 0.1 M NaOH
53
SOLUBILITY EQUILIBRIA
54
Saturated solutions of salts are another type of
chemical equilibria. For a saturated solution of
AgCl, the equation would be AgCl(s) ?Ag(aq)
Cl-(aq) The solubility product expression
would be Ksp AgCl-The AgCl(s)
is left out since solids are left out of
equilibrium expressions (constant
concentrations). For Ag2CO3, Ag2CO3
? 2Ag CO32- Ksp Ag2CO32-
55
The Ksp of AgCl is 1.6 x 10-10. This means that
if the product of AgCl- lt 1.6 x 10-10, the
solution is unsaturated and no solid would be
present. If the product 1.6 x 10-10, the
product is exactly saturated and no solid would
be present. If the product gt 1.6 x 10-10, the
solution is saturated and a solid (precipitate)
would form. The product of the ions (raised to
the power of their coefficients) is called the
ion product constant or Q. If Q gt Ksp, ppt
forms. If Q lt Ksp, no ppt forms.
56
Ex. The molar solubility of silver sulfate is
1.5 x 10-2 mol/L. Calculate the solubility
product of the salt.

• Reaction Ag2SO4(s) ? 2Ag SO42-
• Initial ---- 0
  0
• Change -x 2x
  x
• Equil. ---- 3.0 x 10-2
  1.5 x 10-2
• x 1.5 x 10-2
• Since 1.5 x 10-2 mol/L of Ag2SO4 dissolve,
• 1.5 x 10-2 mol/L of SO42- form and 2(1.5 x 10-2
  mol/L) of Ag form.
• Ksp Ag2SO42- (3.0 x 10-2)2(1.5 x 10-2)
  1.4 x 10-5

Remember that molar solubility is x!
57
Ex. Calculate the molar solubility of calcium
phosphate. The Ksp of calcium phosphate is 1.2 x
10-26.

• Reaction Ca3(PO4)2 ? 3Ca2 2PO43-
• Initial --- 0
  0
• Change -x 3x
  2x
• Equilibrium --- 3x
  2x
• Ksp Ca23PO43-2
• 1.2 x 10-26 (3x)3(2x)2 108x5
• x5 1.1 x 10-28
• x 2.6 x 10-6 M

58
Ex. What is the molar solubility of lead(II)
iodide in a 0.050 M solution of sodium iodide?

• (Common ion effect problem)
• Reaction PbI2 ? Pb2 2I-
• Initial ---- 0 0.050M
  
• Change -x x 2x
• Equil. ---- x 0.050
  2x
• Ksp Pb2I-2
• Ksp 1.4 x 10-8 (x)(0.0502x)2 ? x(0.050)2
  
• x 5.6 x 10-6M
• The molar solubility of PbI2 in pure water is
• 1.5 x 10-3M. This shows the decreased solubility
  of a salt in the presence of a common ion.

Dont forget to put in the initial concentration
of the common ion!
59
Ex. Exactly 200 mL of 0.040 M BaCl2 are added to
exactly 600 mL of 0.080 M K2SO4. Will a
precipitate form?

• BaCl2 K2SO4 ? BaSO4 2KCl Barium sulfate
  is the likely precipitate.
• 0.200 L x 0.040 M BaCl2 8.0 x 10-3 mol Ba2
• 8.0 x 10-3 mol/0.800L total volume 1.0 x 10-2 M
  Ba2
• 0.600 L x 0.080 M K2SO4 4.8 x 10-2 mol SO42-
• 4.8 x 10-2 mol/0.800L 6.0 x 10-2M SO42-
• Ksp Ba2SO42- 1.1 x 10-10 (look this up
  in table)
• Q (1.0 x 10-2)(6.0 x 10-2) 6.0 x 10-4
• Q gt Ksp 6.0 x 10-4 gt 1.1 x 10-10
• A precipitate of BaSO4 forms.